It is known that, as shown in the figure, e and F are two points on the diagonal AC of the parallelogram ABCD, AE = CF

It is known that, as shown in the figure, e and F are two points on the diagonal AC of the parallelogram ABCD, AE = CF

It is proved that: (1) AF = CE and ABCD are parallelograms, AF = CB, ad ‖ BC. In △ ADF and △ CBE, AF = cead = CB, DAF = BCE, AF = CEA, ADF = CB, DAF = BCE, and ≌ ADF ≌ CBE (SAS). (2) AF = ADF ≌ CBE, DFA = bec. DF ≌ EB

As shown in the figure, the planes of two congruent squares ABCD and abef intersect AB, m ∈ AC, n ∈ FB, and am = FN

Proof 1: through M make MP ⊥ BC, NQ ⊥ be, P and Q are perpendicular feet (as shown in the figure), connecting PQ. ∵ MP ∥ AB, NQ ∥ AB, ∥ MP ∥ NQ. And NQ = 22bn = 22cm = MP, ∥ mpqn is parallelogram.. Mn ∥ PQ, PQ ⊂ plane BCE. And Mn ⊄ plane BCE, ∥ Mn ⊄ plane BCE. Proof 2: through M make mg ∥ BC, intersect AB at point G (as shown in the figure), connecting ng ∵ mg ⊂ BC, BC ⊂ plane BCE, Mg ⊄ plane BCE It is also proved that GN ‖ plane BCE, Mg ∩ ng = g, Mn ⊂ plane BCE, Mn ⊂ plane MNG, Mn ⊂ plane BCE

As shown in the figure, the planes of two congruent squares ABCD and abef intersect AB, m ∈ AC, n ∈ FB, and am = FN

Proof 1: through M make MP ⊥ BC, NQ ⊥ be, P and Q are perpendicular feet (as shown in the figure), connecting PQ. ∵ MP ∥ AB, NQ ∥ AB, ∥ MP ∥ NQ. And NQ = 22bn = 22cm = MP, ∥ mpqn is parallelogram.. Mn ∥ PQ, PQ ⊂ plane BCE. And Mn ⊄ plane BCE, ∥ Mn ⊄ plane BCE. Proof 2: through M make mg ∥ BC, intersect AB at point G (as shown in the figure), connecting ng ∵ mg ⊂ BC, BC ⊂ plane BCE, Mg ⊄ plane BCE It is also proved that GN ‖ plane BCE, Mg ∩ ng = g, Mn ⊂ plane BCE, Mn ⊂ plane MNG, Mn ⊂ plane BCE

It is known that the plane of square ABCD and square abef intersects with AB, and the point m.n is on AC and BF respectively, and am = FN

Since it's a geometric problem, let's draw a graph first. Draw a graph according to the idea, and then draw an auxiliary line. Through M, Mg is perpendicular to BC, perpendicular to G; through N, NH is perpendicular to be, perpendicular to h, connecting GH. Because GH is on the plane BCE, we only need to prove that Mn is parallel to GH. Note that these two squares have a common edge, so they are congruent, so AC equals BF, So cm = ac-am = bf-fn = BN. It is easy to know that the triangle CMG is equal to the triangle BNH, so mg = NH, because mg is perpendicular to BC, AB is perpendicular to BC, so mg is parallel to ab. similarly, NH is parallel to AB, so mg and NH are parallel and equal, so mnhg is a parallelogram, so Mn is parallel to GH, so Mn is parallel to plane BCE
P. S: This is only a incomplete proof, because the format and rigor of the proof are still required in the college entrance examination. Think more about it in the improvement. Vector and geometric methods are very important problem-solving methods. If you have the ability, you should be proficient in both